Answer:
[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹
Explanation:
The balanced equation is
H₂O + Cl₂O ⇌ 2HOCl
Data:
Kc = 0.0900
[H₂O] = 0.004 32 mol·L⁻¹
[Cl₂O] = 0.004 32 mol
1. Set up an ICE table.
2. Calculate the equilibrium concentrations
![K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHOCl%5D%24%5E%7B2%7D%24%7D%7D%7B%5Ctext%7B%5BH%24_%7B2%7D%24O%5D%5BCl%24_2%24O%5D%7D%7D%20%3D%20%5Cdfrac%7B%282x%29%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%3D%200.0900%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B4x%5E%7B2%7D%7D%7B%280.00432%20-%20x%29%5E%7B2%7D%7D%20%26%3D%26%200.0900%5C%5C%20%5Cdfrac%7B2x%20%7D%7B0.00432%20-%20x%7D%20%26%20%3D%20%26%200.300%5C%5C2x%20%26%20%3D%20%26%200.300%280.00432%20-%20x%29%5C%5C2x%20%26%20%3D%20%26%200.001296%20-%200.300x%5C%5C2.300x%20%26%20%3D%20%26%200.001296%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B5.63%5Ctimes%2010%5E%7B-4%7D%7D%5C%5C%5Cend%7Barray%7D)
[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹
[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹
Check:
OK.
Answer:
2C3H18 + 15O2 ---->6CO2 +18H2O
Explanation:
the number of reactant must be equal to the number of product.
The density of gasoline is 0.7 g/cm3, and the density of water is 1 g/cm3. Thus the mass of the gasoline is 55*0.7 = 38.5g and the mass of the water is 60g.
Combining the 55 cm3 and 60 cm3 of substances with the aforementioned masses yields a volume of 55+60 = 115cm3 and a mass of 38.5+60 = 98.5g. The density is therefore 98.5/115 = 0.86 g/m3.
C₅H₁₂ + 8O₂ = 5CO₂ + 6H₂O
5CO₂ - 10O
6H₂O - 6O
16O - 8O₂
8 moles of molecular oxygen are consumed
Answer: By carrying out series of wet test
Explanation: Dissolve the ionic compound in water. Ionic compounds will ionise and allow the solution to conduct an electric current. Afterwards specific test will be carried out to detect the presence of cations and anions. Another way would be to find the substance's melting point. Ionically bonded compounds have much higher melting points.
